Example 1: Choose R1 = R2 = R3 = 10 Ω
1. Longer line on battery is
higher voltage and the
shorter line is lower voltage,
i1
by convention.
2. We label the current in
1
each section as shown.
Note that we can guess the ib
direction at this point. The
answer will be independent
of our guess.
ib
It is good to be able to think through (without lots of equations)
how the components of these circuits work.
However, we will need to use our equations to work out
detailed values for current and electric potential in more
complicated circuits.
We now outline a systematic procedure for analyzing circuits.
2 i2
i3
3
ib
ib
+
V=12V
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3. Apply the Voltage Loop
Rule around each possible
loop in the circuit.
Critical Sign Convention!
2 i2
Make sure to label the
direction of your loop!
Make sure to pick a starting
point to go around the loop. ib
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i1
+
V=12V
If your loop goes through a battery from – to +
the Voltage increases (e.g. ΔV = +12 V)
3
1
ib
ib
Then add up the Voltage
changes around the loop.
-
i3
ib
-
+
V=12V
If your loop goes through a battery from + to –
the Voltage decreases (e.g. ΔV = -12 V)
+
V=12V
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i
2
i
2
40
Critical Sign Convention!
Starting at Point P go around the loop.
If you go across a resistor and the loop
direction and guessed current direction are
the same, the voltage decreases (e.g. ΔV = -iR)
ΔVbat = +12V
ΔVR 3 = −i3 R3
ΔVR1 = −i1 R1
If you go across a resistor and the loop
direction and guessed current direction are
opposite, the voltage increases (e.g. ΔV = +iR)
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2 i2
i1
ib 0
ΔVsum = +12 − i3 R3 − i1 R1 =
P
i3
3
1
ib
ib
ib
+
V=12V
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1
For our specific circuit, we used R1 = R2 = R3 = 10 Ω
Now try other path starting at Point P around the new loop.
2 i2
ΔVbat = +12V
ΔVR 2 = −i2 R2
ΔVR1 = −i1 R1
i1
ΔVsum = +12 − i2 R2 − i1 R1 = 0
P
i3
3
1
12 − 10i3 − 10i1 = 0
ΔVsum = +12 − i2 R2 − i1 R1 = 0
12 − 10i2 − 10i1 = 0
We have two equations and three unknowns.
ib
ib
-
ΔVsum = +12 − i3 R3 − i1 R1 = 0
We need additional information.
+
V=12V
ib
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44
12 − 10i3 − 10i1 = 0
Apply the current junction rule at point Q below.
iin = iout
2 i2
i2 + i3 = i1
Also, (as we could have
done at the start):
i1
Q
i2 + i3 = i1
3
1
ib
i1 = ib
12 − 10i2 − 10i1 = 0
i3
Solve 3 equations and 3 unknowns.
ib
ib
-
i2 = 0.4 Amps
i3 = 0.4 Amps
+
V=12V
ib
i1 = 0.8 Amps
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Clicker Question
Does this make sense?
i2 = i3 since two parallel, equal
resistance paths.
Current from R2 and R3 must go
into R1.
i1 = 0.8 Amps
i2 = 0.4 Amps
i3 = 0.4 Amps
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2 i2
i1
i3
3
1
i1
i1
i1
i1
+
V=12V
47
R1 = 10 Ω i1 = 0.8 Amps
R2 = 10 Ω i2 = 0.4 Amps
What is the value of the
equivalent resistor that R3 = 10 Ω i3 = 0.4 Amps
i1
would replace the three
resistors below?
A) Req = 5 Ω
1
B) Req = 10 Ω
i1
C)Req = 15 Ω
D)Req = 30 Ω
E) None of the Above
i1
2 i2
i3
3
i1
i1
+
V=12V
48
2
R23 =
1
⎛ 1
1 ⎞
⎜⎜ R + R ⎟⎟
3 ⎠
⎝ 2
Clicker Question
R1 = 10 Ω i1 = 0.8 Amps
R2 = 10 Ω i2 = 0.4 Amps
R3 = 10 Ω i3 = 0.4 Amps
Two ways to think about it.
1. Battery of 12 V has 0.8 Amps
coming out of it. By V=iR the
Req = 12V/0.8A=15 Ω
2. R2 and R3 are in parallel.
i1
= 5Ω
i3
3
1
i1
Then R23 and R1 are in series.
-
A. a and b
B. a and c
C. b and c
D. a, b, and c
E. a, b, and d
i1
i1
Req = R123 = R1 + R23 = 15Ω
Which of these diagrams represent the same circuit?
2 i2
+
V=12V
i1
49
Clicker Question
50
Clicker Question
A circuit with two batteries is shown below. The directions of
the currents have been chosen (guessed) as shown.
Which is the correct current equation for this circuit?
A) I2 = I1 + I3
B) I1 = I2 + I3
C) I3 = I1 + I2
I1
D) None of these.
Answer: –V2 + I1R1 – I2R2 = 0
R1
V1
R2
Which equation below is the correct equation for Loop 1?
B) V2 + I1R1 – I2R2 = 0
A) –V2 + I1R1 – I2R2 = 0
D) V2 + I1R1 + I2R2 = 0
C) –V2 – I1R1 + I2R2 = 0
E) None of these.
I1
I3
V2
R1
V1
R2
V2
I2 Loop 1.
I2 Loop 1.
R3
R3
I3
I3
51
Clicker Question
Finish a full analysis of this circuit. Consider case:
V1 = 12 V, V2 = 24 V, R1=R2=R3 = 1 Ω.
Which equation below is the correct equation for Loop 1?
B) V2 + I1R1 – I2R2 = 0
A) –V2 + I3R3 – I2R2 = 0
C) –V2 – I1R1 + I2R2 = 0
D) V2 + I1R1 + I2R2 = 0
I1
E) None of these.
I3
Answer: –V2 + I1R1 – I2R2 = 0
Thus none of the above.
Do not try to remember
letter answers.
R2
I1
I3
R1
V1
R2
Loop 2
V2
I2 Loop 1.
R3
R1
V1
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V2
I2 Loop 1.
R3
I3
i1 = 20 A, i2 = −4 A, i3 = 16 A
I3
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i3 = i1 + i2
− V2 + i1 R1 − i2 R2 = 0
− 24 + i1 − i2 = 0
+ V1 − i2 R2 − i3 R3 = 0
+ 12 − i2 − i3 = 0
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3
Clicker Question
Circuit Probes
An Ammeter measures the current through itself.
An ideal Ammeter has zero resistance so it does not affect the
circuit!
Two light bulbs A and B are connected in
series to a constant voltage source. When a
wire is connected across B as shown, the
brightness of bulb A ...
A: increases
B: decreases, but remains glowing
C: decreases to zero (bulb A goes
completely dark, no current)
D: remains unchanged
i
i
A
A
V=5V
I
R
To measure the current
through R, must place
Ammeter in series.
Answer: Bulb A increases in brightness
(since the current increases when B is
shorted).
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Clicker Question
56
Circuit Probes
Consider the circuit shown. If you want to measure the current
thru bulb 3, how should the ammeter be attached (assume you
only attach one at a time)?
A) Pink
Pink
1
B) Yellow
A
A
Blue
A
C) Green
4
3
Yellow 2
V
D) Blue
A
Green
E) Two of the 4 positions
can be used to measure
bulb3’s current.
To measure voltage across R, must place Voltmeter in
parallel with R.
You want the Voltmeter to be very high resistance (ideally
infinity) to avoid drawing current.
V=5V
I
R
V
Answer: B) Yellow
57
Clicker Question
Clicker Question
Now you switch the voltmeter over to "amp" mode. (But you
leave it in the same position in the circuit) What does the
Ammeter read?
A:
6A
B:
3A
6V
A
2Ω
C:
2A
In the circuit shown, what does the voltmeter read?
A:
B:
C:
D:
E:
6V
3V
2V
0V
Voltmeter will "fry"
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6V
2Ω
V
D:
E:
0A
Ammeter will "fry"
Ammeter will fry. Ideal Ammeter has zero internal resistance. If you attach an
ideal ammeter to a battery, you will get infinite power.
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